2000 AMC 10 Problem 20

P16-20~4 minPrint

Algebraic Manipulation Absolute Value & Inequalities Apply an identity: SFFT, sum of squares, difference of cubes, Vieta Consider the largest/smallest element or boundary case

Let $A$ , $M$ , and $C$ be nonnegative integers such that $A+M+C=10$ . What is the maximum value of $A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A$ ?

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Key insight

Adding A + M + C + 1 = 11 turns the expression into (A+1)(M+1)(C+1). Three numbers adding to 13 have the biggest product when nearly equal: 4, 4, 5.

Solution

1. Try some numbers. Call the expression $E = AMC + AM + MC + CA$ .
- $A, M, C = 10, 0, 0$ : $E = 0$ .
- $5, 5, 0$ : $E = 0 + 25 + 0 + 0 = 25$ .
- $3, 3, 4$ : $E = 36 + 9 + 12 + 12 = 69$ .

Numbers close together seem to do best. Now let us prove $69$ is the most.
2. Spot a product. Multiply out $(A+1)(M+1)(C+1)$ :
$(A+1)(M+1)(C+1) = AMC + AM + MC + CA + A + M + C + 1.$
The first four terms are $E$ . The rest is $A + M + C + 1 = 10 + 1 = 11$ . So
$E = (A+1)(M+1)(C+1) - 11.$
3. New question. Now we want three whole numbers $A+1$ , $M+1$ , $C+1$ (each at least $1$ ) that add to $13$ and have the biggest product.
4. Why equal is best. Suppose two of them differ by $2$ or more, like $6$ and $3$ . Moving $1$ from the bigger to the smaller gives $5$ and $4$ . The sum stays the same, but the product goes from $18$ to $20$ . This always happens: moving $1$ from the bigger to the smaller raises the product. So at the maximum, no two of the numbers differ by $2$ or more.
5. Three whole numbers adding to $13$ , all within $1$ of each other, must be $4, 4, 5$ . Their product is $80$ .
6. So the largest $E$ is $80 - 11 = 69$ , reached at $A, M, C = 3, 3, 4$ , as found in step 1.

The answer is $\boxed{\textbf{(C)}\ 69}$ .

Why this works

An expression that looks like "most of" a multiplied-out product usually becomes a product once you add the missing terms. Here the missing terms were exactly $A + M + C + 1$ , which is known. Then a fixed sum with the biggest product means the numbers should be as equal as possible, like a square having the most area of all rectangles with a given perimeter.

Alternative approach

Trust the "as equal as possible" idea and compare nearby choices directly. $3, 3, 4$ gives $69$ ; $2, 4, 4$ gives $32 + 8 + 16 + 8 = 64$ ; $2, 3, 5$ gives $30 + 6 + 15 + 10 = 61$ . The best is $69$ , but the product argument above is what shows no other choice beats it.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The expression looks like most of an expanded product. Indeed,
$(A+1)(M+1)(C+1) = AMC + AM + MC + CA + (A + M + C) + 1,$
and since $A + M + C = 10$ , the target expression equals
$(A+1)(M+1)(C+1) - 11.$

So we maximize a product of three positive integers $a = A+1$ , $b = M+1$ , $c = C+1$ with $a + b + c = 13$ . If two of them differ by $2$ or more, say $a \ge c + 2$ , replacing $(a, c)$ by $(a - 1, c + 1)$ keeps the sum and raises the product, because $(a-1)(c+1) = ac + (a - c) - 1 > ac$ . So at the maximum all three are within $1$ of each other: $\{4, 4, 5\}$ , product $80$ .

Therefore the maximum is $80 - 11 = 69$ , achieved at $(A, M, C) = (3, 3, 4)$ : check $36 + 9 + 12 + 12 = 69$ .

The answer is $\boxed{\textbf{(C)}\ 69}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 11: Special Factorizations (§11.2, §11.3, §11.5)
Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Introduction to Algebra, Chapter 22: Special Manipulations
Introduction to Algebra, Chapter 15: More Inequalities (§15.1, §15.3, §15.4)